What is the CW complex structure of $S^2 × S^2$ which is obtained from $S^2 \vee S^2$ by attaching a $4$-cell?
What I know is that $S^2$ can be given a CW complex structure where the $0$-th skeleton is just a point, $1$-th skeleton is same as that of the $0$-th skeleton and the $2$-th skeleton is obtained from the $1$-th skeleton by attaching a single $2$-cell via the attaching map where we identify the boundary of the $2$-cell to a single point of the $1$-th skeleton. Thus the $2$-th skeleton is a sphere. Then $S^2 \times S^2$ can be given a CW-complex structure where the $n$-th skeleton of $S^2 × S^2$ is obtained by taking the union of the cartesian product of the $p$-th skeleton of $S^2$ with the $q$-th skeleton of $S^2$ over all possible non-negative integers $p$ and $q$ with $p + q = n.$ If we denote the $n$-th skeleton of $S^2 \times S^2$ by $(S^2 \times S^2)^n$ then we find that
$$\begin{align*} (S^2 \times S^2)^0 & = * \times * \\ (S^2 \times S^2)^1 & = * \times * \\ (S^2 \times S^2)^2 & = (S^2 \times *) \cup (* \times S^2) \cup (* \times *) \\ & = (S^2 \times *) \cup (* \times S^2) \\ (S^2 \times S^2)^3 & = (S^2 \times *) \cup (* \times S^2) \\ (S^2 \times S^2)^4 & = S^2 \times S^2 \end{align*}$$
Here our professor claimed that $(S^2 \times S^2)^3 = (S^2 \times S^2)^2 = S^2 \vee S^2$ and thus $S^2 \times S^2$ can be obtained by attaching a $4$-cell to $S^2 \vee S^2.$ But I can't see why? Could anybody please help me in this regard?
Thanks in advance.
Best Answer
If we have $CW$-complexes $X, Y$, then $X \times Y$ also has the structure of a $CW$-complex with $$(X \times Y)^0 = X^0 \times Y^0 \\ (X \times Y)^1 = X^0 \times Y^1 \cup X^1 \times Y^0 \\ (X \times Y)^n = \bigcup_{i=0}^n X^i \times Y^{n-i}$$ $X^{i+1}$ is obtained from $X^i$ by attaching $(i+1)$-cells $D^{i+1}_\alpha$, $\alpha \in A_i$, via maps $\phi_\alpha : \partial D^{i+1}_\alpha \to X^i$, similarly $Y^{j+1}$ is obtained from $Y^j$ by attaching $(j+1)$-cells $D^{j+1}_\beta$, $\beta \in B_j$, via maps $\psi_\beta : \partial D^{j+1}_\beta \to Y^j$. Let the associated characteristic maps be $\bar \phi_\alpha : D^{i+1}_\alpha \to X^{i+1}$ and $\bar \psi_\beta : D^{j+1}_\beta \to Y^{j+1}$.
Then $(X \times Y)^{n+1}$ is obtained from $(X \times Y)^n$ by attaching $(n+1)$-cells $D^{n+1}_{\alpha,\beta} = D^i_\alpha \times D^{n+1-i}_\beta$, $\alpha \in A_i, \beta \in B_{n+1-i}$, $i = 0,\ldots, n+1$. These have characteristic maps $$\bar \chi_{\alpha,\beta} = \bar \phi_\alpha \times \bar \psi_\beta : D^i_\alpha \times D^{n+1-i}_\beta \to X^{i} \times Y^{n+1-i} \subset (X \times Y)^{n+1} .$$ The attaching maps are therefore given by $$\chi_{\alpha,\beta} = \bar \chi_{\alpha,\beta} \mid_{\partial( D^i_\alpha \times D^{n+1-i}_\beta)} : \partial(D^i_\alpha \times D^{n+1-i}_\beta) = \partial (D^i_\alpha) \times D^{n+1-i}_\beta \cup D^i_\alpha \times \partial( D^{n+1-i}_\beta) \to \\ X^{i-1} \times Y^{n+1-i} \cup X^i \times Y^{n-i} \subset (X \times Y)^n .$$ On $\partial (D^i_\alpha) \times D^{n+1-i}_\beta$ the map $\chi_{\alpha,\beta}$ is given by $\phi_\alpha \times \bar \psi_\beta$ and on $D^i_\alpha \times \partial( D^{n+1-i}_\beta)$ by $\bar \phi_\alpha \times \psi_\beta$.
$S^2$ has one $0$-cell and one $2$-cell, hence we have $(S^2)^0 = (S^2)^1 = *, (S^2)^2 = S^2$. Thus $S^2 \times S^2$ has one $0$-cell, two $2$-cells and one $4$-cell. More precisely $$(S^2 \times S^2)^0 = (S^2 \times S^2)^1 = * \times * \\ (S^2 \times S^2)^2 = (S^2 \times S^2)^3 = S^2 \times * \cup * \times S^2 \\ (S^2 \times S^2)^4 = S^2 \times S^2$$ The characteristic maps are given as above.